My question is whether the following variant of the implicit function theorem holds: Let $U \subseteq \mathbb{R}^k$, $V \subseteq \mathbb{R}^m$ and $W \subseteq \mathbb{R}^n$ be open sets and $F$ a $C^1$-function
$$F:U \times V \times W \to \mathbb{R}^{m+n},\phantom{aa}F=F(x,y,z)\phantom{aa}\text{where }x \in U, \ y \in V, \ z \in W$$
such that for some point $(u_0,v_0,w_0) \in U \times V \times W$, we have $F(u_0,v_0,w_0)=0$ and the matrix
$$\frac{\partial F}{\partial(y,z)}$$
is invertible at $(u_0,v_0,w_0)$. Then there are open sets $U' \subseteq U$, $V'\subseteq V$ and $W'\subseteq W$ around $u_0$, $v_0$ and $w_0$ such that we have a unique implicit $C^1$-function
$$g:U'\to V' \times W'$$
with $g(u_0)=(v_0,w_0)$ and $F(x,g(x))=0$ for all $x \in U'$.
I am asking this question since in the classical formulation I know, the domain of $F$ is only a product of two open sets, not of three. That's the point of my question.
Best Answer
Yes, this is just a special case of the usual formulation. To see why, let $X=V\times W$; this is open in $\Bbb{R}^m\times\Bbb{R}^n\cong \Bbb{R}^{m+n}$, and let us use the notation $\xi=(y,z)\in X\subset\Bbb{R}^{m+n}$ and $\xi_0:=(v_0,w_0)$. Then, you have mapping $F:U\times X\subset \Bbb{R}^k\times\Bbb{R}^{m+n}\to\Bbb{R}^{m+n}$ such that by assumption, $F(u_0,\xi_0)=0$ and \begin{align} \frac{\partial F}{\partial \xi}\bigg|_{(u_0,\xi_0)}:\Bbb{R}^{m+n}\to\Bbb{R}^{m+n} \end{align} is an invertible linear transformation/matrix. Hence, there exist open $\tilde{U}\subset U$ about $u_0$, open $\tilde{X}\subset X$ about $\xi_0$ and an implicit $C^1$ solution $g:\tilde{U}\to\tilde{X}$ such that $g(u_0)=\xi_0$ and for all $x\in \tilde{U}$, we have $F(x,g(x))=0$.
Now, with some basic topology we can get the formulation you're after. By how the product topology works, we can find open $V'\subset V$ about $v_0$, open $W'\subset W$ about $w_0$ such that $V'\times W'\subset\tilde{X}$. Finally, define $U':= g^{-1}(V'\times W')$. Then, the restricted mapping $g:U'\to V'\times W'$ is still such that $g(u_0)=\xi_0:=(v_0,w_0)$, and such that for all $x\in U'$, we have $F(x,g(x))=0$.